Let $X$ and $Y$ be random variables such that $f$ is the joint density of $(X,Y)$.
If $\int_{-\infty}^{\infty}f(x,y)dx>0$ for every $y\in\mathbb{R}$ and $g$ is a measurable function such that $E\lvert g(X)\rvert<\infty$, one can easily check that \begin{equation} E(g(X)\mid Y)=h(Y) \end{equation} where $h(y):=\int_{-\infty}^{\infty}g(x)f(x,y)\,dx/\int_{-\infty}^{\infty}f(x,y)\,dx$. According to Rick Durrett's book we can get rid of the hypothesis of $\int_{-\infty}^{\infty}f(x,y)\,dx$ being strictly positive just by defining $h$ as we please when $\int_{-\infty}^{\infty}f(x,y)\,dx=0$, but I have been struggling to prove this without success. Any idea?
Just in case it helps, I leave here the proof of the easy part:
It is clear that $h(Y)\in\sigma(Y)$, then, Let $A\in\sigma(Y)$, so $A=Y^{-1}(B)$, being $B\subset\mathbb{R}$ some borelian set. \begin{equation} \int_Ah(Y)=\int_Bh(y)\int_{-\infty}^{\infty}f(x,y)\,dx\,dy=\int_B\int_{-\infty}^{\infty}g(x)f(x,y)\,dx\,dy \end{equation}
\begin{equation} \int_Ag(X)=\int_\Omega g(X)1_A=\int_{\Omega}j(X,Y)=\int_{B}\int_{-\infty}^{\infty}g(x)f(x,y)\,dx\,dy \end{equation}